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Aptitude Questions: LCM & HCF Set 4

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Hello Aspirants. Welcome to Online Quantitative Aptitude Section with explanation in AffairsCloud.com. Here we are creating question sample in LCM & HCF, which is common for all the IBPS, SBI exam and other competitive exams. We have included questions that are repeatedly asked in bank exams !!!

  1. What is the HCF of 3/5, 7/10, 2/15, 6/21?
    A) 5/123
    B) 6/121
    C) 2/105
    D) 1/210
    E) None of these
    D) 1/210
    Explanation:

    HCF will be HCF of numerators/LCM of denominators
    So HCF = HCF of 3,7,2,6)/(LCM of 5,10, 15, 21)
    = 1/210

  2. The HCF and LCM of two numbers is 84 and 840 respectively. If the first number is 168, find the second one.
    A) 210
    B) 420
    C) 225
    D) 326
    E) None of these
    B) 420
    Explanation:

    Product of two numbers = HCF * LCM
    So 2nd number = 84*840/168

  3. Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs of numbers will satisfy this condition?
    A) 2
    B) 5
    C) 3
    D) 6
    E) 4
    E) 4
    Explanation:

    Since HCF = 8 is highest common factor among those numbers
    So, let first no = 8x, 2nd number = 8y
    So 8x + 8y = 128
    x + y = 16
    the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9)
    *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.
    Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)

  4. Find the least number exactly divisible by 10, 15, 18 and 30.
    A) 85
    B) 90
    C) 88
    D) 93
    E) 105
    B) 90
    Explanation:

    It will be the LCM of these numbers.

  5. Find the least number which when divided by15, 21, 24 and 32 leaves the same remainder 2 in each case.
    A) 3362
    B) 3360
    C) 3456
    D) 3266
    E) 3364
    A) 3362
    Explanation:

    It will be the LCM of these numbers + 2
    So 3360 + 2

  6. Find the least number which when divided by 10, 15, 18 and 30 leaves remainders 6, 11, 14 and 26 respectively.
    A) 72
    B) 86
    C) 85
    D) 90
    E) 94
    B) 86
    Explanation:

    Since 10-6 = 4, 15-11 = 4, 18-14 = 4, 30-26 = 4
    So answer will be the LCM of these numbers – 4
    So 90 – 4

  7. Find the smallest number of 4 digits which is exactly divisible by 12, 15, 21 and 30.
    A) 1580
    B) 1420
    C) 1260
    D) 1056
    E) None of these
    C) 1260
    Explanation:

    Least 4 digit number = 1000
    LCM of (12, 15, 21, 30) = 420
    On dividing 420 by 1000, leaves remainder = 160
    So answer = 1000 + (420 – 160)

  8. Find the least number which when divided by 2, 5, 9 and 12 leaves a remainder 3 but leaves no remainder when same number is divided by 11.
    A) 281
    B) 357
    C) 360
    D) 363
    E) 224
    D) 363
    Explanation:

    LCM(2, 5, 9, 12) = 180
    So the number will be somewhat = 180x + 3
    This number leaves no remainder when divided by 11, so x = 2 to get (180x+3) fully divided by 11
    So number = 180*2 + 3

  9. HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers.
    A) 15/242
    B) 16/275
    C) 32/221
    D) 20/268
    E) None of these
    B) 16/275
    Explanation:

    x + y = 80, xy = 5*275 = 1375
    sum of reciprocals = 1/x + 1/y = (x+y)/xy = 80/1375

  10. The HCF and LCM of two numbers is 70 and 1050 respectively. If the first number is 210, find the second one.
    A) 450
    B) 350
    C) 510
    D) 260
    E) None of these
    B) 350
    Explanation:

    Produvt of two numbers = HCF * LCM
    So 2nd number = 70*1050/210